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Distance Rate and Time

Mathematics is a vast subject which has many branches. Algebra is one of the most important and main branches of mathematics. It is the study of problems based on constants, variables, equations and expressions. It deals with symbols and rules of manipulation of symbols. There are various mathematical branches which are connected with algebra, such as - number theory, calculus, geometry, arithmetic etc.

Algebra itself has many subdivisions. Elementary algebra includes the study of elementary operations with constants and variables. It includes techniques of solving elementary algebraic equations. Modern or abstract algebra is the advanced field which studies about the abstract parts of mathematics. Algebra is very important for having the knowledge of any other science-related subject, such as - engineering and technology, physics, chemistry, even economics and medicine.

The concept of study about the distance, speed (rate) and time of something is very commonly used in algebra. We come across with various problems based upon distance, rate and time in our day-to-day life. Let us go ahead and understand this concept and problems related to this.

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The distance is the length of space traveled by a moving object. It may also be defined as the length measured between two points. The distance is usually denoted by "d". It is a scalar quantity which means it has only magnitude, but no direction. The unit of distance is that of length. It is usually measured in kilometer (km), meter (m), centimeter (cm), millimeter (mm).

In algebraic calculations, the time is a measure of an event to be happen. In distance, rate, time problems, the time is measured as the fraction in which a particular distance is travelled. The time is eventually a scalar quantity. It is generally denoted by "t". Its standard unit is "second". However, it sometimes appear in hours and minutes, but it should be converted into seconds before calculations.

In algebra, the rate is also known as speed. It is defined as the amount of distance travelled in the unit time, i.e. distance per unit time. It is also a scalar quantity and is denoted by "r" or "s". The rate is measured in meter per second, kilometer per hour, or miles per hour.


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1) The formula for the rate or speed of an object is explained below:
$r$ = $\frac{d}{t}$

Where, r denotes rate or speed, d represents distance travelled and t denotes the time taken in covering the given distance.
$Rate$ = $\frac{Distance}{Time}$
2) The formula for the distance can be derived from the above formula:
Distance = Rate x Time
d = r x t
3) Similarly, the formula for time is shown below:
$Time$ = $\frac{Distance}{Rate}$
$t$ = $\frac{d}{r}$


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The following instructions should be kept in mind while solving problems based on time, distance and rate:
1) Firstly, read the question carefully and identify the quantities (usually any two of distance, rate or time should be given) involved in the question.

2) Have a look at the units of given quantities. They must be similar. If they are not, convert and make them similar. i.e. if rate is given in meter per second, then time and distance should be measured in seconds and meter respectively. In the same way, if the rate is given km/hr, then distance and time are measured in kilometer and hour respectively.

3) Now, by using the three formulae discussed in above section, we may calculate the third quantity with the help of two given quantities.

Word Problems

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Few word problems based on the concepts of distance, rate and time are illustrated below:

Problem 1: James travels from one city to another.  The time taken in covering whole trip is 3.5 hours, when he traveled at an average rate of 62 miles/hour . How many miles does James travel ?

Solution: Speed or rate = 62 miles/hour, Time = 3.5 hours, Distance = ?

We have the formula :

Distance = Rate x Time

Distance = 62 x 3.5 217 miles.

Problem 2: Bob completed a bicycle trip in 4 hours. He covered a distance of 84 kilometer. Find the average rate in meter/second.

Solution: Time = 4 hours = 4 x 3600 seconds

Distance = 84 km = 84000 m

Rate = $\frac{Distance}{Time}$

Rate = $\frac{84000}{4 \times 3600}$

Rate = 5.8 m/s

Problem 3: A train crosses the a man standing on platform in 18 seconds and the platform in 30 seconds when it travels at an average speed of 72 km/h. Calculate the length of the platform in meters?

Solution: The distance covered by the train when it crosses the man should be equal to the length of the train

Also, the distance covered by the train when it crosses the platform is equal to the sum of length of the train and length of platform.

30 seconds - 18 seconds = 12 seconds which is the extra time taken by the train while crossing the platform on account of length of the platform.

Speed or rate = 72 km/h = $\frac{72 \times 1000}{3600}$ m/s = 20 m/s.
The extra time which is taken in covering the platform = 12 seconds

Length of platform = rate x extra time

Length of platform = 20 x 12 = 240 meters.
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